Natural log & L'Hopitals $x^x$

Question

I need someone to tell me the step I'm missing or doing incorrectly. The problem is:

$$\lim _{ x\rightarrow 0 }{ { x }^{ x } } $$

$1$. $\ln x^x$

$2$. $x\ln { x } $

$3$. This is the step I don't follow: $\frac { \ln { x } }{ \frac { 1 }{ x } } $

I come up with: $\frac { \ln { x } }{ x } $ at this point.

I'm missing this step in all the natural $\log$/l'hopitals problems I work. I picked the most simple for ease of explanation, if someone could be so kind?

Answer 1

$$\lim _{ x\rightarrow 0 }{ { x }^{ x }= } \lim _{ x\rightarrow 0 }{ { e }^{ x\ln { x } }= } \lim _{ x\rightarrow 0 }{ { e }^{ \frac { \ln { x } }{ \frac { 1 }{ x } } }\overset { L'h\quad rule }{ = } } \lim _{ x\rightarrow 0 }{ { e }^{ \frac { \frac { 1 }{ x } }{ -\frac { 1 }{ { x }^{ 2 } } } }= } \lim _{ x\rightarrow 0 }{ { e }^{ -x }=1 } $$

Answer 2

Even without L'Hospital, considering $$A=x^x \implies \log(A)=x \log(x)\implies \log(A)\to 0\implies A \to 1$$

Answer 3

I interpret your question to be about how $$x\ln x =\frac{\ln x}{\frac1x}$$

Here, they're using the rule that $\displaystyle x = \frac1{\frac1x}$ and that $\frac1b \cdot a = \frac ab$

That means that, with a bit more steps, we have $$x\cdot \ln x = \frac1{\frac1x}\cdot \ln x = \frac{\ln x}{\frac1x}$$

The reason this is useful, is that the limit now gives an indeterminate form that qualifies for L'Hopital.